Answer
See step by step solution.
Work Step by Step
Let's look at a triangle $\triangle ABC$ and the line with equation $y=mx+b$
$$\tan\theta = \dfrac{BC}{CA}.$$
Let's find coordinates of points A,B,C.
$A_{y}=0$ and as A is on the line $y=mx+b$. Hence $x=-\dfrac{b}{m}$ and $A_{x}=-\dfrac{b}{m}$. $A(-b/m,0)$
For the point C we just have $C_{x}$ is some arbitrary x and $C(x,0)$
For the point B we have the same x coordinate with C -- $x$. And as B is also on the line $y=mx+b$ we get $B(x,mx+b)$.
Now we can find sides BC and CA.
$$BC=\sqrt{(B_{x}-C_{x})^2+(B_{y}-C_{y})^2}=\sqrt{(x-x)^2+(mx+b-0)^2}=mx+b.$$
Here we consider that $mx+b$ is greater than zero as shown in the picture.
$$CA=\sqrt{(C_{x}-A_{x})^2+(C_{y}-A_{y})^2}=\sqrt{(-b/m-x)^2+(0-0)^2}=\sqrt{(-1)^2(b/m+x)^2}.$$
$$CA=b/m+x$$
We also consider that $b/m+x$ is greater than zero.
Now we have
$$\tan\theta=\dfrac{mx+b}{\dfrac{b}{m}+x}=\dfrac{m(mx+b)}{b+mx}=m$$
